SPOJ
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REPEATS - Repeats
A string s is called an (k,l)-repeat if s is obtained by concatenating k>=1 times some seed string t with length l>=1. For example, the string
s = abaabaabaaba
is a (4,3)-repeat with t = aba as its seed string. That is, the seed string t is 3 characters long, and the whole string s is obtained by repeating t 4 times.
Write a program for the following task: Your program is given a long string u consisting of characters ‘a’ and/or ‘b’ as input. Your program must find some (k,l)-repeat that occurs as substring within u with k as large as possible. For example, the input string
u = babbabaabaabaabab
contains the underlined (4,3)-repeat s starting at position 5. Since u contains no other contiguous substring with more than 4 repeats, your program must output the maximum k.
Input
In the first line of the input contains H- the number of test cases (H <= 20). H test cases follow. First line of each test cases is n - length of the input string (n <= 50000), The next n lines contain the input string, one character (either ‘a’ or ‘b’) per line, in order.
Output
For each test cases, you should write exactly one interger k in a line - the repeat count that is maximized.
Example
Input:117babbabaabaabaababOutput:4since a (4, 3)-repeat is found starting at the 5th character of the input string.
往后各能匹配到多远,记这个总长度为K,那么这里连续出现了K/L+1次。最后看最大值是多少。
于是,我们不妨先算一下,从r[i*j]开始,除匹配了k/i个重复子串,还剩余了几个字符,剩余的自然是k%i个字符。如果说r[i*j]的前面还有i-k%i个字符完成匹配的话,这样就相当于利用多余的字符还可以再匹配出一个重复子串,于是我们只要检查一下从r[i*j-(i-k%i)]和r[i*(j+1)-(i-k%i)]开始是否有i-k%i个字符能够完成匹配即可,也就是说去检查这两个后缀的最长公共前缀是否比i-k%i大即可。
当然如果公共前缀不比i-k%i小,自然就不比i小,因为后面的字符都是已经匹配上的,所以为了方便编写,程序里面就直接去看是否会比i小就可以了。
/*Sherlock and Watson and Adler*/#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<list>#include<bitset>#include<cmath>#include<complex>#include<string>#include<algorithm>#include<iostream>#define eps 1e-9#define LL long long#define PI acos(-1.0)#define bitnum(a) __builtin_popcount(a)using namespace std;const int N = 5005;const int M = 100005;const int inf = 1000000007;const int mod = 1000000007;const int MAXN = 50005;//rnk从0开始//sa从1开始,因为最后一个字符(最小的)排在第0位//height从1开始,因为表示的是sa[i - 1]和sa[i]//倍增算法 O(nlogn)int wa[MAXN], wb[MAXN], wv[MAXN], ws_[MAXN];//Suffix函数的参数m代表字符串中字符的取值范围,是基数排序的一个参数,如果原序列都是字母可以直接取128,如果原序列本身都是整数的话,则m可以取比最大的整数大1的值//待排序的字符串放在r数组中,从r[0]到r[n-1],长度为n//为了方便比较大小,可以在字符串后面添加一个字符,这个字符没有在前面的字符中出现过,而且比前面的字符都要小//同上,为了函数操作的方便,约定除r[n-1]外所有的r[i]都大于0,r[n-1]=0//函数结束后,结果放在sa数组中,从sa[0]到sa[n-1]void Suffix(int *r, int *sa, int n, int m){ int i, j, k, *x = wa, *y = wb, *t; //对长度为1的字符串排序 //一般来说,在字符串的题目中,r的最大值不会很大,所以这里使用了基数排序 //如果r的最大值很大,那么把这段代码改成快速排序 for(i = 0; i < m; ++i) ws_[i] = 0; for(i = 0; i < n; ++i) ws_[x[i] = r[i]]++;//统计字符的个数 for(i = 1; i < m; ++i) ws_[i] += ws_[i - 1];//统计不大于字符i的字符个数 for(i = n - 1; i >= 0; --i) sa[--ws_[x[i]]] = i;//计算字符排名 //基数排序 //x数组保存的值相当于是rank值 for(j = 1, k = 1; k < n; j *= 2, m = k) { //j是当前字符串的长度,数组y保存的是对第二关键字排序的结果 //第二关键字排序 for(k = 0, i = n - j; i < n; ++i) y[k++] = i;//第二关键字为0的排在前面 for(i = 0; i < n; ++i) if(sa[i] >= j) y[k++] = sa[i] - j;//长度为j的子串sa[i]应该是长度为2 * j的子串sa[i] - j的后缀(第二关键字),对所有的长度为2 * j的子串根据第二关键字来排序 for(i = 0; i < n; ++i) wv[i] = x[y[i]];//提取第一关键字 //按第一关键字排序 (原理同对长度为1的字符串排序) for(i = 0; i < m; ++i) ws_[i] = 0; for(i = 0; i < n; ++i) ws_[wv[i]]++; for(i = 1; i < m; ++i) ws_[i] += ws_[i - 1]; for(i = n - 1; i >= 0; --i) sa[--ws_[wv[i]]] = y[i];//按第一关键字,计算出了长度为2 * j的子串排名情况 //此时数组x是长度为j的子串的排名情况,数组y仍是根据第二关键字排序后的结果 //计算长度为2 * j的子串的排名情况,保存到数组x t = x; x = y; y = t; for(x[sa[0]] = 0, i = k = 1; i < n; ++i) x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j]) ? k - 1 : k++; //若长度为2 * j的子串sa[i]与sa[i - 1]完全相同,则他们有相同的排名 }}int Rank[MAXN], height[MAXN], sa[MAXN], r[MAXN];void calheight(int *r,int *sa,int n){ int i,j,k=0; for(i=1; i<=n; i++)Rank[sa[i]]=i; for(i=0; i<n; height[Rank[i++]]=k) for(k?k--:0,j=sa[Rank[i]-1]; r[i+k]==r[j+k]; k++);}int n,minnum[MAXN][16];void RMQ() //预处理 O(nlogn){ int i,j; int m=(int)(log(n*1.0)/log(2.0)); for(i=1;i<=n;i++) minnum[i][0]=height[i]; for(j=1;j<=m;j++) for(i=1;i+(1<<j)-1<=n;i++) minnum[i][j]=min(minnum[i][j-1],minnum[i+(1<<(j-1))][j-1]);}int Ask_MIN(int a,int b) //O(1){ int k=int(log(b-a+1.0)/log(2.0)); return min(minnum[a][k],minnum[b-(1<<k)+1][k]);}int calprefix(int a,int b){ a=Rank[a],b=Rank[b]; if(a>b) swap(a,b); return Ask_MIN(a+1,b);}char s[5];int main(){ int t,i,j,k,ans,Max; scanf("%d",&t); while(t--) { Max=1; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%s",s); r[i]=s[0]-'a'+1; } r[i]=0; Suffix(r,sa,n+1,3); calheight(r,sa,n); RMQ(); //枚举重复长度 for(i=1;i<=n;i++) { //找到每一段 for(j=0;j+i<n;j+=i) { //先求出前缀 ans=calprefix(j,j+i); k=j-(i-ans%i); ans=ans/i+1; if(k>=0&&calprefix(k,k+i)>=i) ans++; //printf("L=%d,R=%d\n",i,ans); Max=max(Max,ans); } } printf("%d\n",Max); } return 0;}
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